bmm10234 chapter 4 - circles.pdf

7/28/2019 BMM10234 Chapter 4 - Circles.pdf

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Chapter 4 | Circles | BMM10234 | 92 of 204

Arc of a circle

Pat f the cicumfeence f a cicle is called an ac

A B

Q

P

MAJOR ARC

MINOR ARC

Chord of a circle

The line segment jining any tw pints n the cicumfeence is called a chd f the cicle The chd unless it is als a

diamete divides the cicle int tw acs min ac and maj ac

The lage ac is the maj ac (AP) The shte ac is the min ac (AQ)

The diamete can als be defined as the chd passing thugh the cente f the cicle

The lagest chd f the cicle is the diamete

The lnge chd f the cicle is neae t the cente than the shte chd

Tw acs int which a cicle is divided by a chd ae called cnjugate acs

Relation between chord and arc

(a) Unequal chds cut ff unequal min acs and unequal maj acs

(b) Of the tw unequal chds the geate chd cuts ff a geate min ac than the smalle ne des

(c) Of the tw unequal chds the geate chd cuts ff a smalle maj ac than the smalle ne des

(d) Equal chds cut - ff equal maj acs and equal min acs

Segments of a circle

MINOR SEGMENT

MAJOR SEGMENT

O

A B

Q

P

A als divides the cicle int tw pats Each pat is called a segment f a cicle

A segment f a cicle can be defined as a plane figue bunded by a chd n ne side and an ac n the the side The

chd is the base f the segment

The segment bunded by the chd and the maj ac is called the maj segment (AP) The cente f the cicle lies

in the maj segment

The segment bunded by the chd and the min ac is called the min segmentA diamete divides the cicle int tw equal segments

Each segment is called the semi-cicle

A min segment is less than a semi-cicle and a maj segment is geate than a semicicle


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O is the cente f the cicle A is a chd in it then AP = P

Cnvesely the line jining the mid-pint f any chd f a cicle t the cente is pependicula t the chd

Equal chds f a cicle ae equidistant fm the cente

O

D

P

A C

B

Q

If A = C then OP = OQ

Cnvesely the chds which ae equidistant fm the cente ae equal

If OP = OQ then A = C

Solved ExamplesExample 1:

Find the length f a chd which is at a distance f cms fm the cente f a cicle whse adius is 5 cms

A P B

O

4 cm

Solution:

Let AP = P = I then A = I

OA = OP is the distance f the chd fm the centeGiven OA = 5 cm A = I cms

y Pythagas theem

OA= OP

+ AP

(OP is pependicula t A)

5=

+ AP

5 = AP

(5 + ) (5 ) = AP

9 = AP AP = I = A = I = = cms

Example 2:

In a cicle f adius 5 cms A and C ae tw paallel chds f length 8 cms and cms espectively Calculate the distance

between the chds if they ae:

(i) On the same side f the cente(ii) On the ppsite sides f the cente


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Example 3:

In the adjining figue equal chds A and C cut at ight angles at X M and N ae the mid pints f A and C Pve the

OMXN is a squae

A B

O

M X

C

D

N

Solution:

In the quadilateal OMXN

O + + + =

O + 9 + 9 + 9 =

OM is pependicula t A = 9

ON is pependicula t C = 9 = 9 (given)O = 7 = 9

Theefe in the quadilateal all the angles ae ight angles OMXN can be cnsideed a ectangleON = MX ()

OM = NX () (ppsite sides f a ectangle)

ut OM = ON () (Equal chds ae equidistant fm the cente)

Fm () () and ()

OM = ON = MX = NX

If in a ectangle all the sides ae equal then it is a squae

OMXN is a squae

Example 4:

A ectangle with ne side equal t cms is inscibed in a cicle f adius 5 cms Find the aea f the ectangle

A B

O

CD

4 cm

FE

Solution:

The cente f the cicle is the pint f intesectin f the diagnals f the ectangle aw OE pependicula t A

EO pduce meets C at F

OF is pependicula t C

EF || A (OEA = 9and A = 9

)

EF = A OE = OF

OE =1

2A =

1

2 = cms

In ight angled tiangle OE


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=

2r

2

+

22 r

r4

=

2 2 2 2

2 24r r 3r 3r BD or BD BD4 4 4

= = =

=3.r

2 = 3.r

C = 3.r

Theefe the length f the side f an equilateal tiangle inscibed in a cicle f adius is 3r

Example 7:

If a staight line APQ is dawn t cut tw cncentic cicles pve that AP = Q

AP

B

O

D Q

Solution:

aw O pependicula t A which is the chd f the bigge cicle and PQ is the chd f smalle cicle O is pependicula

t A and PQ

() A = and () P = Q (pependicula fm the cente bisects the chd)

() () gives A P = Q AP = Q

Sector of a circle

A B

D

O

r r

C

The ptin f cicle enclsed between tw adii and an ac is called sect In the abve figue the sect OA is the min

sect epesented by the shaded egin The sect OCA is the maj sect If Ac A subtends an angle at the cente

then the aea f the sect2r

,360

=

o

whee is the adius f the cicle


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aw a cicle with cente O Mak an ac PRQ Jin PO and QO POQ is the angle subtended at the cente by the ac PRQMak a pint M n the ac PAQ Jin PM and QM PMQ is the angle subtended by the same ac at the cicumfeence

Theorem

In a cicle the angle subtended by an ac at the cente is duble the angle subtended by the same ac at the cicumfeence

OC = AC

A

O

B C

Angles in the same segment ae equal

B

A

P Q R

O

The angle in a semi-cicle is a ight angle

BA

Q

O

The angle in a maj segment is an acute angle and the angle in the min segment is an btuse angle

BA

Q

O

P

Acute

Obtuse

Theorem

Oppsite angles f a cyclic quadilateal ae supplementay

(A quadilateal whse vetices lie n the cicumfeence f a cicle is called a cyclic quadilateal)


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CAO = O (angles in the same segment)Theefe tiangle AOC is simila t tiangle O (Tiangle ae equi - angula)

AO CO

DO BO= (sides f simila tiangle ae pptinal)

AO O = COO (by css multiplicatin)

Example 12:

C is a chd f a cicle with cente O A is a pint n an ac as shwn in the figue Pve that

(a) AC + OC = 9 if A is a pint n the maj ac(b) AC OC = 9 if A is a pint n the min ac

A

O

B Cy y

x

Solution:(a) OC = x (angle made by the ac in the cente is twice the angle n the cicumfeence)

In tiangle OC

OC + y + y = 8

OC = 8 y x + y = 8

(x + y) = 8 x + y = 9 AC + OC = 9 (x = AC y = OC)

(b)

A

O

B C

180

y

z

yx

D

C = 8 x (ppsite angle f cyclic quadilateal)& OC = C OC = xIn OC OC + y + y = 8 OC = 8 y 8 = x y x y = 9

Example 13:

Given a cyclic tapezium AC in which A ||C and C = 7 detemine its the angles

B

A

O

C

D

700


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AE = AC = 5

AE = 5 AE = 5

A > AE (extei angle is geate than the intei ppsite angles in a tiangle)A > 5

Example 16:

In the figue O is the cente f a cicle YQ = 95

and QY =

Find PQ PAQ and POQ

AB

Y

O20

0

950

P Q

Solution:

In tiangle YQ 95+ + = 8

= 8 5 = 5

PQ = PAQ (angles in the same segment)PAQ = 5

POQ = (angle in the cente is twice the angle at the cicumfeence)POQ = 5 =

Tangent of a circle

A B

O

P

A line which meets a cicle at ne pint nly which when pduced des nt cut it again is called a tangent In the figue A

is the tangent The pint P is called the pint f cntact

Secant

A BC D

O

Any line dawn fm a pint utside the cicle and meeting the cicle at tw pints is called a secant A is a secant which

meets the cicle at C and


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Case

(i) If d = R + the cicles tuch extenally

(ii) If d = R the cicles tuch intenally (R > )

(iii) If d > R + the cicles d nt tuch

(iv) If d < R + the cicles intesect

(v) If d = the cicles ae cncentic

Angle in an alternate segment

A

B

P Q

YX

O

A is a chd f the cicle with cente O A divides the cicle int tw segment AX and AY PAQ is tangent dawn t the

cicle at A AQ and AP ae the angles fmed by the chd A with the tangent PAQ AQ and segment AX ae nppsite sides f the chd A The segment AX is called the altenate with efeence t angle AQ Similaly segmentAY is the altenate segment with efeence t angle AP

Theorem

In a cicle the angles between a tangent and the chd at the pint f cntact ae espectively equal t the angles in the

altenate segments

A

BP

Q

YX

O

XAY is tangent at A t the cicle with cente O; A is a chd If P is any pint n the altenate segment with efeence t

AY and Q is any pint in the altenate segment with espect t AX then it can be pved that (i) AY = AP (ii)AX = AQ

Common tangents to two circles

(i) If d = R + thee cmmn tangents can be dawn t the cicle The tangents ae A C and XY A and C ae diect

cmmn tangents XY is the tansvese cmmn tangent Length f cmmn tangent A &C is given by t = 2 2d (R r)

A B

DY

X

C

C1

C2


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Theorem:

If PA secant t a cicle intesects the cicle at A and and PT is a tangent then PA P = PT

B

A

P

T

Theorem:

If tw chds f a cicle intesect inside utside the cicle when pduced then the fllwing elatin hlds

B

A

P

C

O

P

A

D

D

C B

O

A and C ae tw chds which intesect at P (in figue () inside the cicle in fig () utside the cicle) then PA P = PC

P

Example 17:

In tw cncentic cicles pve that a chd f lage cicle which is tangent t a smalle cicle is bisected at the pint fcntact

O

QBA

Solution:

Let A be the chd f the lage cicle tuching the smalle cicle at Q Jin OQ Nw OQ is the adius f the smalle cicle

OQ = 9

Again A is the chd f the bigge cicle OQ is pependicula t A

Theefe AQ = Q


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Solution:

OP pependicula t PT (The tangent is pependicula t the adius)

OPT = 9

In OPT OPT + POT + T = 8

9+ POT + = 8

POT = 8 =

Example 21:

O is the cente f the cicle PA and P ae tangents If PA = 7 find PA and AP

O

P

B

A

700

Solution:

PA = P (Tangents dawn fm an extenal pint ae equal)

PA = PA = 7 PA = 7

AP = 8 (PA + PA)=8 (7 + 7)AP = 8 = AP =

Example 22:

AP is a tangent at P t the cicle with cente O If angle QP = find the angle POQ

O

Q

PA

600

B

Solution:

AP is a tangent and OP is the adius OP = 9

QP =

OPQ = OP QPOPQ = 9 =

In OPQ OP = OQ (adius)OPQ = OQP =

POQ = 8 ( + ) =

POQ =

Example 23:

Thee cicles with centes A and C tuch each the extenally at P Q and R If A = 5 cms C = 7 cms and CA = cms find

the adius f each cicle


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Example 25:

In the figue AX = 5 Find the angles f the tiangle AC if A = AC

AX Y

B C

500

Solution:

A = AC (given)

AX = AC = 5 (angle in the altenate segment)AC = AC = 5

AC = 5

AC = 8 (5 + 5) = 8 = 8

AC = 8 AC = 5, AC = 5

Example 26:

Tw cicles intesect at A and PQ is a cmmn tangent Shw that PAQ + PQ = 8

A

B

C1

C2

P Q

Solution:F the cicle with the cente C

P is the chd and PQ is the tangent

PQ = PA (Angle made by the tangent with chd is equal t angle in the altenate segment)LetPQ = PA = xF the cicle with cente C

QP = QA= yAdding () and ()

PQ + QP = PA + QAPQ + QP = x + yIn PQPQ + QP + PQ = 8

PQ = 8 (x + y) [because PQ + QP = x + y]PAQ = PA + QAPAQ = x + y

PAQ + PQ = 8 (x + y) + (x + y)PAQ + PQ = 8


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Thus Aea f the segment = A A' = (8 99) = 59 cm

Example 29:

AC is a tiangle in which A = cms and AC = 5 cms C is the diamete f the cicle thugh and C Find the aea f the

cicle

A

BCO

5 12

Solution:

C is the diameteAC = 9 (angle in the semi cicle)C

= A

+ AC

(Pythagas theem)

= + 5

= + 5 = 9

C = cms

iamete = C = cmsRadius = 5 cms

Aea f the cicle =2 22r 6.5 6.5 1 32.7 sq. cms (approx )

7 = =

Example 30:

If AP = P and PA = find AP AQ and AO

BA

PQ

O

Solution:

In tiangle AP AP = P

PA = PA =

In tiangle AP AP + PA + PA = 8

AP + + =8

AP = 8 =

AP = AQ = (angle in the same segment)AQ =

AO = AP = = AO =


13/17


In the given figue oAOC 120 . = Find m CBE , whee O is the cente: (Access Cde - )

() () () () 5 (5) Nne f these

5 In the given figue AC is a cyclic quadilateal and A is athe diameteoADC 140 , = then find m BAC: (Access

Cde - 5)

() 5 () () 5 () (5) Nne f these

In the given figue O is the cente f the cicleoAOB 90 . = Find m APB: (Access Cde - )

()

() 5

() 5

() 55

(5) Cannt be detemined


14/17


In the diagam if pints A C ae pints f tangency the x equals: (Access Cde - )

()3

cm16

()1

cm8

()1

cm32

()3

cm32

(5)1

cm16

Fm pint P utside a cicle with a cicumfeence f units a tangent is dawn Als fm P a secant is dawn dividing the

cicle int unequal acs with lengths m and n It is fund that t the length f the tangent is the mean pptinal betweenm and n If m and t ae integes then t may have the fllwing numbe f values: (Access Cde - )

() ze () ne () tw () thee (5) infinitely many

5 In the figue B is equal t (Access Cde - 5)

() 85 () 95 () 7 () 5 (5) Nne f these

The sun f the adii f the tw cicles is cm and diffeence between thei cicumfeence is 88 cm The adius f the lage

cicle is (Access Cde - )

() cm () 7 cm () cm () 77 cm (5) 8 cm

7 A squae is inscibed is a semi cicle f adius cms What is the aea f the squae inscibed given a side f the squae is

alng the diamete f the semicicle? (Access Cde - 7)

() 8 cm () 5 cm () 5 cm () cm (5) 7 cm

8 Tw cncentic cicles with cente C have adii and

such that

2 1(r r ) 0. CA > and CA ae the cmmn lined adii f the

cicles If tangent at A is dawn t meet the bigge cicle in the pint The the length is given by (Access Cde -8)

()1 2 12r (r r ) () 2 2 12r (r r ) () 1 2 12r (r r )+ () 2 2 12r (r r )+ (5) Nne f these


15/17


5 A is a fixed diamete f a cicle whse cente is O Fm C any pint n the cicle a chd C is dawn pependicula t A

Then as C mves ve a semicicle the bisect f angle OC cuts the cicle in a pint that always: (Access Cde - 5)

() bisects the ac A

() tisects the ac A

() vaies

() is as fa fm A as fm

(5) is squidistant fm and c

Use HB pencil only. Abide by the time-limit

SCORE SHEET

5

5

5

5

5

5

7

8

9

5

5

5

5

5

5

5

5

5

5

5

7

8

9

5

5

5

5

5

5

5

5

5

5

5


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() r 2

()

() Nt a side f an incibed egula plygn

()r 3

2

(5) r 3

In this figue the cente f the cicle is O A C AOE is a staight line AP = AD, and A has a lenght twice the adiusThen: (Access Cde - 9)

() AP = PB. AB2 () AP. DO = PB. AD () AB = AD. DE

2 () AB. = AD = OB. AO (5) Nne f these

5 Cnside a hmbus with diagnals f length and 8 and the inscibed cicle C The vetices f the hmbus ae the

midpints f the sides f a ectangle inscibed in a cicle C The ati f the adius f C

and the adius f C

is (Access

Cde - )

() less than 8 () 8 () 5 () (5) Geate than

Fm pint P utside a cicle with a cicumfeence f units a tangent is dawn Als fm P a secant is dawn dividing the

cicle int unequal acs with lengths m and n It is fund that t the length f the tangent is the mean pptinal between

m and n If m and t ae integes then t may have the fllwing numbe f values: (Access Cde - )

() ze () ne () tw () thee (5) infinitely many

7 iamete A f a cicle with cente O is units C is a pint units fm A and n A is a pint units fm and nA P is any pint n the cicle Then the bken-line path fm C t P t : (Access Cde - )

() has the same length f all psitins f P

() exceeds units f all psitins f P

() cant exceed units

() is shtest when CP is a ight tiangle

(5) is lngest when P is equidistant fm C and

8 In cicle O G is a mving pint n diamete A AA is dawn pependicula t A and equal t AG is dawn pependicu-

la t A n the same side f diamete A as AA and equal tBG Let O be the midpint f A Then as G mves fm

A t pint O (Access Cde - )

() mves n a staight line paallel t A

() emains statinay() mves n a staight line pependicula t A

() mves in a small cicle intesectin the given cicle

(5) fllws a path which is neithe a cicle n a staight line


17/17


(5)

()

(5)

()

5 ()

()

()

()

()

5 ()

()

7 ()

8 ()

9 ()

()

()

()

(5)

()

5 ()

()

7 ()

8 ()

9 ()

(5)

Practice Exercise - 1

Answers

Practice Exercise - 2

(5)

()

()

()

5 ()

( )

7 (5)

8 ()

9 ()

()

bmm10234 chapter 4 - circles.pdf

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